Tài liệu Oc klein 2e 2014 1373p ssm_

Student Study Guide and Solutions Manual Organic Chemistry Second Edition David Klein           Stu nt Study Guide  uden and lutio Man , 2e  d So ons M nual,     for   Organic C mistry, Chem , 2e    Davi id Klein  Johns s Hopkin ns Unive ersity                                    This book is printed on acid free paper.        Founded in 1807, John Wiley & Sons, Inc. has been a valued source of knowledge and understanding for more than  200 years, helping people around the world meet their needs and fulfill their aspirations. Our company is built on a  foundation of principles that include responsibility to the communities we serve and where we live and work. In  2008, we launched a Corporate Citizenship Initiative, a global effort to address the environmental, social,  economic, and ethical challenges we face in our business. Among the issues we are addressing are carbon impact,  paper specifications and procurement, ethical conduct within our business and among our vendors, and  community and charitable support. For more information, please visit our  website:  www.wiley.com/go/citizenship.    Copyright  2015, 2012    John Wiley & Sons, Inc.  All rights reserved.  No part of this publication may be  reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical,  photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United  States Copyright Act, without either the prior written permission of the Publisher, or authorization through  payment of the appropriate per‐copy fee to the Copyright Clearance Center, Inc. 222 Rosewood Drive, Danvers,  MA 01923, website www.copyright.com.  Requests to the Publisher for permission should be addressed to the  Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken,  NJ 07030‐5774, (201)748‐6011, fax  (201)748‐6008, website http://www.wiley.com/go/permissions.      Evaluation copies are provided to qualified academics and professionals for review purposes only, for use in their  courses during the next academic year.  These copies are licensed and may not be sold or transferred to a third  party.  Upon completion of the review period, please return the evaluation copy to Wiley.  Return instructions and  a free of charge return shipping label are available at www.wiley.com/go/return label. Outside of the United  States, please contact your local representative.      Main: ISBN 978‐1‐118‐64795‐0  Binder Version: ISBN 978‐1‐118‐70081‐5      Printed in the United States of America    10  9  8  7  6  5  4  3  2  1  CONTENTS    Chapter 1 – Electrons, Bonds, and Molecular Properties       1  Chapter 2 – Molecular Representations       31  Chapter 3 – Acids and Bases       69  Chapter 4 – Alkanes and Cycloalkanes       100  Chapter 5 – Stereoisomerism       130  Chapter 6 – Chemical Reactivity and Mechanisms       155  Chapter 7 – Substitution Reactions       178  Chapter 8 – Alkenes:  Structure and Preparation via Elimination Reactions       213  Chapter 9 – Addition Reactions of Alkenes       256   Chapter 10 – Alkynes       297  Chapter 11 – Radical Reactions       339  Chapter 12 – Synthesis       374  Chapter 13 – Alcohols and Phenols       407  Chapter 14 – Ethers and Epoxides; Thiols and Sulfides       456  Chapter 15 – Infrared Spectroscopy and Mass Spectrometry       502  Chapter 16 – Nuclear Magnetic Resonance Spectroscopy       531  Chapter 17 – Conjugated Pi Systems and Pericyclic Reactions       572  Chapter 18 – Aromatic Compounds       610  Chapter 19 –Aromatic Substitution Reactions       642  Chapter 20 – Aldehydes and Ketones       707  Chapter 21 – Carboxylic Acids and Their Derivatives       776  Chapter 22 – Alpha Carbon Chemistry: Enols and Enolates       836  Chapter 23 – Amines       912  Chapter 24 – Carbohydrates       969  Chapter 25 – Amino Acids, Peptides, and Proteins       994  Chapter 26 – Lipids        1017  Chapter 27 – Synthetic Polymers        1032  HOW TO USE THIS BOOK    Organic chemistry is much like bicycle riding.  You cannot learn how to ride a bike by watching  other people ride bikes.  Some people might fool themselves into believing that it’s possible to  become an expert bike rider without ever getting on a bike.  But you know that to be incorrect  (and very naïve).  In order to learn how to ride a bike, you must be willing to get on the bike,  and you must be willing to fall.  With time (and dedication), you can quickly train yourself to  avoid falling, and to ride the bike with ease and confidence.  The same is true of organic  chemistry.  In order to become proficient at solving problems, you must “ride the bike”.  You  must try to solve the problems yourself (without the solutions manual open in front of you).   Once you have solved the problems, this book will allow you to check your solutions.  If,  however, you don’t attempt to solve each problem on your own, and instead, you read the  problem statement and then immediately read the solution, you are only hurting yourself.  You  are not learning how to avoid falling.  Many students make this mistake every year.  They use  the solutions manual as a crutch, and then they never really attempt to solve the problems on  their own.  It really is like believing that you can become an expert bike rider by watching  hundreds of people riding bikes.  The world doesn’t work that way!    The textbook has thousands of problems to solve.  Each of these problems should be viewed as  an opportunity to develop your problem‐solving skills.  By reading a problem statement and  then reading the solution immediately (without trying to solve the problem yourself), you are  robbing yourself of the opportunity provided by the problem.  If you repeat that poor study  habit too many times, you will not learn how to solve problems on your own, and you will not  get the grade that you want.      Why do so many students adopt this bad habit (of using the solutions manual too liberally)?   The answer is simple.  Students often wait until a day or two before the exam, and then they  spend all night cramming.  Sound familiar?  Unfortunately, organic chemistry is the type of  course where cramming is insufficient, because you need time in order to ride the bike yourself.   You need time to think about each problem until you have developed a solution on your own.   For some problems, it might take days before you think of a solution.  This process is critical for  learning this subject.  Make sure to allot time every day for studying organic chemistry, and use  this book to check your solutions.  This book has also been designed to serve as a study guide,  as described below.      WHAT’S IN THIS BOOK      This book contains more than just solutions to all of the problems in the textbook.  Each  chapter of this book also contains a series of exercises that will help you review the concepts,  skills and reactions presented in the corresponding chapter of the textbook.  These exercises  are designed to serve as study tools that can help you identify your weak areas.   Each chapter  of this solutions manual/study guide has the following parts:     Review of Concepts.  These exercises are designed to help you identify which concepts  are the least familiar to you.  Each section contains sentences with missing words  (blanks).  Your job is to fill in the blanks, demonstrating mastery of the concepts.  To  verify that your answers are correct, you can open your textbook to the end of the  corresponding chapter, where you will find a section entitled Review of Concepts and  Vocabulary.  In that section, you will find each of the sentences, verbatim.     Review of Skills.  These exercises are designed to help you identify which skills are the  least familiar to you. Each section contains exercises in which you must demonstrate  mastery of the skills developed in the SkillBuilders of the corresponding textbook  chapter.  To verify that your answers are correct, you can open your textbook to the end  of the corresponding chapter, where you will find a section entitled SkillBuilder Review.   In that section, you will find the answers to each of these exercises.     Review of Reactions.  These exercises are designed to help you identify which reagents  are not at your fingertips.  Each section contains exercises in which you must  demonstrate familiarity with the reactions covered in the textbook.  Your job is to fill in  the reagents necessary to achieve each reaction.  To verify that your answers are  correct, you can open your textbook to the end of the corresponding chapter, where  you will find a section entitled Review of Reactions.  In that section, you will find the  answers to each of these exercises.     Common Mistakes to Avoid.  This is a new feature to this edition.  The most common  student mistakes are described, so that you can avoid them when solving problems.     A List of Useful Reagents.  This is a new feature to this edition.  This list provides a  review of the reagents that appear in each chapter, as well as a description of how each  reagent is used.   Solutions. At the end of each chapter, you’ll find detailed solutions to all problems in the  textbook, including all SkillBuilders, conceptual checkpoints, additional problems,  integrated problems, and challenge problems.      The sections described above have been designed to serve as useful tools as you study and  learn organic chemistry.  Good luck!      David Klein  Senior Lecturer, Department of Chemistry  Johns Hopkins University              Chapter 1 A Review of General Chemistry: Electrons, Bonds and Molecular Properties Review of Concepts Fill in the blanks below. To verify that your answers are correct, look in your textbook at the end of Chapter 1. Each of the sentences below appears verbatim in the section entitled Review of Concepts and Vocabulary.            _____________ isomers share the same molecular formula but have different connectivity of atoms and different physical properties. Second-row elements generally obey the _______ rule, bonding to achieve noble gas electron configuration. A pair of unshared electrons is called a ______________. A formal charge occurs when an atom does not exhibit the appropriate number of ___________________________. An atomic orbital is a region of space associated with ____________________, while a molecular orbital is a region of space associated with _______________. Methane’s tetrahedral geometry can be explained using four degenerate _____-hybridized orbitals to achieve its four single bonds. Ethylene’s planar geometry can be explained using three degenerate _____-hybridized orbitals. Acetylene’s linear geometry is achieved via _____-hybridized carbon atoms. The geometry of small compounds can be predicted using valence shell electron pair repulsion (VSEPR) theory, which focuses on the number of  bonds and _______________ exhibited by each atom. The physical properties of compounds are determined by __________________ forces, the attractive forces between molecules. London dispersion forces result from the interaction between transient __________________ and are stronger for larger alkanes due to their larger surface area and ability to accommodate more interactions. Review of Skills Fill in the blanks and empty boxes below. To verify that your answers are correct, look in your textbook at the end of Chapter 1. The answers appear in the section entitled SkillBuilder Review. SkillBuilder 1.1 Determining the Constitution of Small Molecules 2 CHAPTER 1 SkillBuilder 1.2 Drawing the Lewis Dot Structure of an Atom SkillBuilder 1.3 Drawing the Lewis Structure of a Small Molecule SkillBuilder 1.4 Calculating Formal Charge SkillBuilder 1.5 Locating Partial Charges Resulting from Induction SkillBuilder 1.6 Identifying Electron Configurations SkillBuilder 1.7 Identifying Hybridization States 3 CHAPTER 1 SkillBuilder 1.8 Predicting Geometry SkillBuilder 1.9 Identifying the Presence of Molecular Dipole Moments SkillBuilder 1.10 Predicting Physical Properties Dipole-Dipole Interactions H-Bonding Interactions Carbon Skeleton CIRCLE THE COMPOUND BELOW THAT IS EX PECTED TO HAVE THE HIGHER BOILING POINT CIRCLE THE COMPOUND BELOW THAT IS EXPECTED TO HAVE THE HIGHER BOILING POINT CIRCLE THE COMPOUND BELOW THAT IS EX PECTED TO HAVE THE HIGHER BOILING POINT CH2 H3C C CH3 H3C H O C CH3 H C H H O C H H H H H C C H H H O H H H H C C C H H H H H H H H H H C C C C C H H H H H H A Common Mistake to Avoid When drawing a structure, don’t forget to draw formal charges, as forgetting to do so is a common error. If a formal charge is present, it MUST be drawn. For example, in the following case, the nitrogen atom bears a positive charge, so the charge must be drawn: As we progress though the course, we will see structures of increasing complexity. If formal charges are present, failure to draw them constitutes an error, and must be scrupulously avoided. If you have trouble drawing formal charges, go back and master that skill. You can’t go on without it. Don’t make the mistake of underestimating the importance of being able to draw formal charges with confidence. 4 CHAPTER 1 Solutions 1.1. (a) Begin by determining the valency of each atom in the compound. The carbon atom is tetravalent, the oxygen atom is divalent, and the hydrogen atoms are all monovalent. The atoms with more than one bond (in this case, C and O) should be drawn in the center of the compound. The hydrogen atoms are then placed at the periphery, as shown. (b) Begin by determining the valency of each atom in the compound. The carbon atom is tetravalent, while the chlorine and hydrogen atoms are all monovalent. The carbon atom is the only atom with more than one bond, so it must be drawn in the center of the compound. The chlorine atom and the hydrogen atoms are then placed at the periphery, as shown. The chlorine atom can be placed in any one of the four available positions. The following four drawings all represent the same compound, in which the central carbon atom is connected to the Cl. (c) The carbon atoms are tetravalent, while the hydrogen atoms are all monovalent. The atoms with more than one bond (in this case, the two carbon atoms) should be drawn in the center of the compound. The hydrogen atoms are then placed at the periphery, as shown. (d) The carbon atom is tetravalent, the nitrogen atom is trivalent, and the hydrogen atoms are all monovalent. The atoms with more than one bond (in this case, C and N) should be drawn in the center of the compound. The hydrogen atoms are then placed at the periphery, as shown. (e) The carbon atoms are tetravalent, while the fluorine atoms are all monovalent. The atoms with more than one bond (in this case, the two carbon atoms) should be drawn in the center of the compound. The fluorine atoms are then placed at the periphery, as shown. (f) The carbon atoms are tetravalent, while the bromine atom and hydrogen atoms are all monovalent. The atoms with more than one bond (in this case, the two carbon atoms) should be drawn in the center of the compound. The bromine atom and hydrogen atoms are then placed at the periphery, as shown. The bromine atom can be placed in any one of the six available positions. The following six drawings all represent the same compound, in which the two carbon atoms are connected to each other, and the bromine atom is connected to one of the carbon atoms. (g) The carbon atoms are tetravalent, while the hydrogen atoms are all monovalent. The atoms with more than one bond (in this case, the three carbon atoms) should be drawn in the center of the compound. The hydrogen atoms are then placed at the periphery, as shown. 1.2. Begin by determining the valency of each atom that appears in the molecular formula. The carbon atoms are tetravalent, while the chlorine atom and hydrogen atoms are all monovalent. The atoms with more than one bond (in this case, the three carbon atoms) should be drawn in the center of the compound. Then, as indicated in the problem statement, the chlorine atom can be placed in either of two locations: i) connected to the central carbon atom, or ii) connected to one of the other two 5 CHAPTER 1 (equivalent) carbon atoms. The hydrogen atoms are then placed at the periphery. 1.3. The carbon atoms are tetravalent, the oxygen atom is divalent, and the hydrogen atoms are all monovalent. The atoms with more than one bond (in this case, C, C, C and O) should be drawn in the center of the compound. There are several different ways to connect these four atoms. We can connect the three carbon atoms in a linear chain (C-C-C), which gives two different locations where the oxygen atom can be placed (see the solution to Problem 1.2). Specifically, the oxygen atom can either be connected to the central carbon atom (C2) or connected to one of the other two carbon atoms (C1 or C3). four carbon atoms and one oxygen atom. Let’s begin with the four carbon atoms. There are two different ways to connect four carbon atoms. They can either be arranged in a linear fashion or in a branched fashion. Next, the oxygen atom must be inserted. For each of the two skeletons above (linear or branched), there are several different locations to insert the oxygen atom. The linear skeleton has four possibilities, shown here: and the branched skeleton has three possibilities shown here: Connecting the oxygen atom to C3 is the same as connecting it to C1 (if we just assign numbers in the other direction, right-to-left). There is another way in which the three carbon atoms and the one oxygen atom can be connected. Instead of connecting the carbon atoms to each other in a chain of three carbon atoms, we can insert the oxygen atom between two of the carbon atoms, like this: Finally, we complete all of the structures by drawing the bonds to hydrogen atoms. H H Inserting the oxygen atom between C2 and C3 is the same as inserting it between C1 and C2 (if we just assign numbers in the other direction, right-to-left). In summary, we have seen exactly three different ways to connect three carbon atoms and one oxygen atom. For each of these possibilities, shown below, the hydrogen atoms are placed at the periphery. O H H H C C C C H H H H H H H H C C C H H C O H H H H H O H H C C C C H H H H H H H C C H H H C H C H H H H H H H C O H C C H H H H H C C O H H H H H H H C O C C H H H 1.4. Begin by determining the valency of each atom that appears in the molecular formula. The carbon atoms are tetravalent, the oxygen atom is divalent, and the hydrogen atoms are all monovalent. Any atoms with more than one bond (in this case, the four carbon atoms and the one oxygen atom) should be drawn in the center of the compound, with the hydrogen atoms at the periphery. There are several different ways to connect H C H C C H C H H H H H H H O H H H H 1.5. (a) Carbon belongs to group 4A of the periodic table, and it therefore has four valence electrons. The periodic symbol for carbon (C) is drawn, and each valence 6 CHAPTER 1 electron is placed by itself (unpaired), around the C, like this: (b) Oxygen belongs to group 6A of the periodic table, and it therefore has six valence electrons. The periodic symbol for oxygen (O) is drawn, and each valence electron is placed by itself (unpaired) on a side of the O, until all four sides are occupied. That takes care of four of the six electrons, leaving just two more electrons to draw. Each of the two remaining electrons is then paired up with an electron already drawn, like this: O (c) Fluorine belongs to group 7A of the periodic table, and it therefore has seven valence electrons. The periodic symbol for fluorine (F) is drawn, and each valence electron is placed by itself (unpaired) on a side of the F, until all four sides are occupied. That takes care of four of the seven electrons, leaving three more electrons to draw. Each of the three remaining electrons is then paired up with an electron already drawn, like this: (d) Hydrogen belongs to group 1A of the periodic table, and it therefore has one valence electron. The periodic symbol for hydrogen (H) is drawn, and the one and only valence electron is placed on a side of the H, like this: (e) Bromine belongs to group 7A of the periodic table, and it therefore has seven valence electrons. The periodic symbol for bromine (Br) is drawn, and each valence electron is placed by itself (unpaired) on a side of the Br, until all four sides are occupied. That takes care of four of the seven electrons, leaving three more electrons to draw. Each of the three remaining electrons is then paired up with an electron already drawn, like this: (f) Sulfur belongs to group 6A of the periodic table, and it therefore has six valence electrons. The periodic symbol for sulfur (S) is drawn, and each valence electron is placed by itself (unpaired) on a side of the S, until all four sides are occupied. That takes care of four of the six electrons, leaving just two more electrons to draw. Each of the two remaining electrons is then paired up with an electron already drawn, like this: (g) Chlorine belongs to group 7A of the periodic table, and it therefore has seven valence electrons. The periodic symbol for chlorine (Cl) is drawn, and each valence electron is placed by itself (unpaired) on a side of the Cl, until all four sides are occupied. That takes care of four of the seven electrons, leaving three more electrons to draw. Each of the three remaining electrons is then paired up with an electron already drawn, like this: (h) Iodine belongs to group 7A of the periodic table, and it therefore has seven valence electrons. The periodic symbol for iodine (I) is drawn, and each valence electron is placed by itself (unpaired) on a side of the I, until all four sides are occupied. That takes care of four of the seven electrons, leaving three more electrons to draw. Each of the three remaining electrons is then paired up with an electron already drawn, like this: 1.6. Both nitrogen and phosphorus belong to group 5A of the periodic table, and therefore, each of these atoms has five valence electrons. In order to achieve an octet, we expect each of these elements to form three bonds. 1.7. Aluminum is directly beneath boron on the periodic table (group 3A), and therefore both elements exhibit three valence electrons. 1.8. The Lewis dot structure for a carbon atom is shown in the solution to Problem 1.5a. That drawing must be modified by removing one electron, resulting in a formal positive charge, as shown below. This resembles boron because it exhibits three valence electrons. 1.9. The Lewis dot structure for a carbon atom is shown in the solution to Problem 1.5a. That drawing must be modified by drawing one additional electron, resulting in a formal negative charge, as shown below. This resembles nitrogen because it exhibits five valence electrons. 1.10. (a) Each carbon atom has four valence electrons, and each hydrogen atom has one valence electron. Only the carbon atoms can form more than one bond, so we begin by connecting the carbon atoms to each other. Then, we connect all of the hydrogen atoms, as shown. CHAPTER 1 (b) Each carbon atom has four valence electrons, and each hydrogen atom has one valence electron. Only the carbon atoms can form more than one bond, so we begin by connecting the carbon atoms to each other. Then, we connect all of the hydrogen atoms, and the unpaired electrons are shared to give a double bond. In this way, each of the carbon atoms achieves an octet. (c) Each carbon atom has four valence electrons, and each hydrogen atom has one valence electron. Only the carbon atoms can form more than one bond, so we begin by connecting the carbon atoms to each other. Then, we connect all of the hydrogen atoms, and the unpaired electrons are shared to give a triple bond. In this way, each of the carbon atoms achieves an octet. (d) Each carbon atom has four valence electrons, and each hydrogen atom has one valence electron. Only the carbon atoms can form more than one bond, so we begin by connecting the carbon atoms to each other. Then, we connect all of the hydrogen atoms, as shown. 1.12. Each of the carbon atoms has four valence electrons; the nitrogen atom has five valence electrons; and each of the hydrogen atoms has one valence electron. We begin by connecting the atoms that have more than one bond (in this case, the three carbon atoms and the nitrogen atom). There are four different ways that these four atoms can be connected to each other, shown below. For each of these possible arrangements, we connect the hydrogen atoms, giving the following four constitutional isomers. H H H H N C C C H H H H H H H H H C N C C H H H H H (e) Each carbon atom has four valence electrons, and each hydrogen atom has one valence electron. Only the carbon atoms can form more than one bond, so we begin by connecting the carbon atoms to each other. Then, we connect all of the hydrogen atoms, and the unpaired electrons are shared to give a double bond. In this way, each of the carbon atoms achieves an octet. (f) The carbon atom has four valence electrons, the oxygen atom has six valence electrons, and each hydrogen atom has one valence electron. Only the carbon atom and the oxygen atom can form more than one bond, so we begin by connecting them to each other. Then, we connect all of the hydrogen atoms, as shown. 1.11. Boron has three valence electrons, each of which is shared with a hydrogen atom, shown below. The central boron atom lacks an octet of electrons, and it is therefore very unstable and reactive. 7 H H C H HNH C H H C H H H H H C N C H HHCH H H In each of these four structures, the nitrogen atom has one lone pair. 1.13. (a) Aluminum is in group 3A of the periodic table, and it should therefore have three valence electrons. In this case, the aluminum atom exhibits four valence electrons (one for each bond). With one extra electron, this aluminum atom will bear a negative charge. (b) Oxygen is in group 6A of the periodic table, and it should therefore have six valence electrons. In this case, the oxygen atom exhibits only five valence electrons (one for each bond, and two for the lone pair). This oxygen atom is missing an electron, and it therefore bears a positive charge. (c) Nitrogen is in group 5A of the periodic table, and it should therefore have five valence electrons. In this case, the nitrogen atom exhibits six valence electrons (one for each bond and two for each lone pair). With 8 CHAPTER 1 one extra electron, this nitrogen atom will bear a negative charge. electron, so this chlorine atom will bear a positive charge. (d) Oxygen is in group 6A of the periodic table, and it should therefore have six valence electrons. In this case, the oxygen atom exhibits only five valence electrons (one for each bond, and two for the lone pair). This oxygen atom is missing an electron, and it therefore bears a positive charge. (i) Two of the atoms in this structure exhibit a formal charge because each of these atoms does not exhibit the appropriate number of valence electrons. The nitrogen atom (group 5A) should have five valence electrons, but it exhibits four (one for each bond). It is missing one electron, so this nitrogen atom will bear a positive charge. One of the two oxygen atoms (the one on the right) exhibits seven valence electrons (one for the bond, and two for each lone pair), although it should have only six. With one extra electron, this oxygen atom will bear a negative charge. (e) Carbon is in group 4A of the periodic table, and it should therefore have four valence electrons. In this case, the carbon atom exhibits five valence electrons (one for each bond and two for the lone pair). With one extra electron, this carbon atom will bear a negative charge. H C H H (f) Carbon is in group 4A of the periodic table, and it should therefore have four valence electrons. In this case, the carbon atom exhibits only three valence electrons (one for each bond). This carbon atom is missing an electron, and it therefore bears a positive charge. (g) Oxygen is in group 6A of the periodic table, and it should therefore have six valence electrons. In this case, the oxygen atom exhibits only five valence electrons (one for each bond, and two for the lone pair). This oxygen atom is missing an electron, and it therefore bears a positive charge. (h) Two of the atoms in this structure exhibit a formal charge because each of these atoms does not exhibit the appropriate number of valence electrons. The aluminum atom (group 3A) should have three valence electrons, but it exhibits four (one for each bond). With one extra electron, this aluminum atom will bear a negative charge. The neighboring chlorine atom (to the right) should have seven valence electrons, but it exhibits only six (one for each bond and two for each lone pair). It is missing one 1.14. (a) The boron atom in this case exhibits four valence electrons (one for each bond), although boron (group 3A) should only have three valence electrons. With one extra electron, this boron atom bears a negative charge. (b) The nitrogen atom in this case exhibits six valence electrons (one for each bond and two for each lone pair). But nitrogen (group 5A) should only have five valence electrons. With one extra electron, this nitrogen atom bears a negative charge. H N H (c) One of the carbon atoms (below right) exhibits three valence electrons (one for each bond), but carbon (group 4A) is supposed to have four valence electrons. It is missing one electron, so this carbon atom therefore bears a positive charge. 1.15. (a) Oxygen is more electronegative than carbon, and a C–O bond is polar covalent. For each C–O bond, the O will be electron rich (‒), and the C will be electron-poor (+), as shown below. CHAPTER 1 (b) Fluorine is more electronegative than carbon, and a C–F bond is polar covalent. For a C–F bond, the F will be electron-rich (‒), and the C will be electron-poor (+). Chlorine is also more electronegative than carbon, so a C–Cl bond is also polar covalent. For a C–Cl bond, the Cl will be electron-rich (‒), and the C will be electron-poor (+), as shown below. 9 bond, the Cl will be electron-rich (‒) and the C will be electron-poor (+), as shown below. Notice that two carbon atoms are electron-poor (+). These are the positions that are most likely to be attacked by an anion, such as hydroxide. (c) Carbon is more electronegative than magnesium, so the C will be electron-rich (‒) in a C–Mg bond, and the Mg will be electron-poor (+). Also, bromine is more electronegative than magnesium. So in a Mg–Br bond, the Br will be electron-rich (‒), and the Mg will be electron-poor (+), as shown below. (d) Oxygen is more electronegative than carbon or hydrogen, so all C–O bonds and all O–H bond are polar covalent. For each C–O bond and each O–H bond, the O will be electron-rich (‒), and the C or H will be electron-poor (+), as shown below. (e) Oxygen is more electronegative than carbon. As such, the O will be electron-rich (‒) and the C will be electron-poor (+) in a C=O bond, as shown below. (f) Chlorine is more electronegative than carbon. As such, for each C–Cl bond, the Cl will be electron-rich (‒) and the C will be electron-poor (+), as shown below. 1.16. Oxygen is more electronegative than carbon. As such, the O will be electron-rich (‒) and the C will be electron-poor (+) in a C=O bond. In addition, chlorine is more electronegative than carbon. So for a C–Cl 1.17. (a) As indicated in Figure 1.10, carbon has two 1s electrons, two 2s electrons, and two 2p electrons. This information is represented by the following electron configuration: 1s22s22p2 (b) As indicated in Figure 1.10, oxygen has two 1s electrons, two 2s electrons, and four 2p electrons. This information is represented by the following electron configuration: 1s22s22p4 (c) As indicated in Figure 1.10, boron has two 1s electrons, two 2s electrons, and one 2p electron. This information is represented by the following electron configuration: 1s22s22p1 (d) As indicated in Figure 1.10, fluorine has two 1s electrons, two 2s electrons, and five 2p electrons. This information is represented by the following electron configuration: 1s22s22p5 (e) Sodium has two 1s electrons, two 2s electrons, six 2p electrons, and one 3s electron. This information is represented by the following electron configuration: 1s22s22p63s1 (f) Aluminum has two 1s electrons, two 2s electrons, six 2p electrons, two 3s electrons, and one 3p electron. This information is represented by the following electron configuration: 1s22s22p63s23p1 1.18. (a) The electron configuration of a carbon atom is 1s22s22p2 (see the solution to Problem 1.17a). However, if a carbon atom bears a negative charge, then it must have one extra electron, so the electron configuration should be as follows: 1s22s22p3 (b) The electron configuration of a carbon atom is 1s22s22p2 (see the solution to Problem 1.17a). However, if a carbon atom bears a positive charge, then it must be missing an electron, so the electron configuration should be as follows: 1s22s22p1 (c) As seen in Skillbuilder 1.6, the electron configuration of a nitrogen atom is 1s22s22p3. However, if a nitrogen atom bears a positive charge, then it must be missing an electron, so the electron configuration should be as follows: 1s22s22p2 (d) The electron configuration of an oxygen atom is 1s22s22p4 (see the solution to Problem 1.17b). However, if an oxygen atom bears a negative charge, then it must have one extra electron, so the electron configuration should be as follows: 1s22s22p5 10 CHAPTER 1 1.19. The bond angles of an equilateral triangle are 60º, but each bond angle of cyclopropane is supposed to be 109.5º. Therefore, each bond angle is severely strained, causing an increase in energy. This form of strain, called ring strain, will be discussed in Chapter 4. The ring strain associated with a three-membered ring is greater than the ring strain of larger rings, because larger rings do not require bond angles of 60º. 1.20. (a) The C=O bond of formaldehyde is comprised of one  bond and one  bond. (b) Each C‒H bond is formed from the interaction between an sp2 hybridized orbital from carbon and an s orbital from hydrogen. (c) The oxygen atom is sp2 hybridized, so the lone pairs occupy sp2 hybridized orbitals. 1.21. Rotation of a single bond does not cause a reduction in the extent of orbital overlap, because the orbital overlap occurs on the bond axis. In contrast, rotation of a  bond results in a reduction in the extent of orbital overlap, because the orbital overlap is NOT on the bond axis. 1.22. (a) The highlighted carbon atom (below) has four bonds, and is therefore sp3 hybridized. The other carbon atoms in this structure are all sp2 hybridized, because each of them has three bonds and one  bond. (b) Each of the highlighted carbon atoms (below) has four bonds, and is therefore sp3 hybridized. The other carbon atoms in this structure are all sp2 hybridized, because each of them has three bonds and one  bond. atoms (the outer ones) are sp2 hybridized because each has three bonds and one  bond. (b) One of the carbon atoms (the one connected to oxygen) has two bonds and two  bonds, and as such, it is sp hybridized. The other carbon atom is sp2 hybridized because it has three bonds and one  bond. 1.24. Carbon-carbon triple bonds generally have a shorter bond length than carbon-carbon double bonds, which are generally shorter than carbon-carbon single bonds (see Table 1.2). 1.25. (a) The nitrogen atom has three bonds and one lone pair, so the steric number is 4 (sp3 hybridization), which means that the electronic arrangement will be tetrahedral. One corner of the tetrahedron is occupied by a lone pair, so the arrangement of atoms is trigonal pyramidal. (b) The oxygen atom has three bonds and one lone pair, so the steric number is 4 (sp3 hybridization), which means that the electronic arrangement will be tetrahedral. One corner of the tetrahedron is occupied by a lone pair, so the arrangement of atoms is trigonal pyramidal. 1.23. (a) Each of the two central carbon atoms has two bonds and two  bonds, and as such, each of these carbon atoms is sp hybridized. The other two carbon (c) The boron atom has four bonds and no lone pairs, so the steric number is 4 (sp3 hybridization), which means that the electronic arrangement will be tetrahedral. With no lone pairs, the arrangement of atoms (geometry) CHAPTER 1 is the same as the electronic arrangement. tetrahedral. It is (d) The boron atom has three bonds and no lone pairs, so the steric number is 3 (sp2 hybridization), which means that the electronic arrangement will be trigonal planar. With no lone pairs, the arrangement of atoms (geometry) is the same as the electronic arrangement. It is trigonal planar. (e) The boron atom has four bonds and no lone pairs, so the steric number is 4 (sp3 hybridization), which means that the electronic arrangement will be tetrahedral. With no lone pairs, the arrangement of atoms (geometry) is the same as the electronic arrangement. It is tetrahedral. (f) The carbon atom has four bonds and no lone pairs, so the steric number is 4 (sp3 hybridization), which means that the electronic arrangement will be tetrahedral. With no lone pairs, the arrangement of atoms (geometry) is the same as the electronic arrangement. It is tetrahedral. (g) The carbon atom has four bonds and no lone pairs, so the steric number is 4 (sp3 hybridization), which means that the electronic arrangement will be tetrahedral. With no lone pairs, the arrangement of atoms (geometry) is the same as the electronic arrangement. It is tetrahedral. (h) The carbon atom has four bonds and no lone pairs, so the steric number is 4 (sp3 hybridization), which means that the electronic arrangement will be tetrahedral. With no lone pairs, the arrangement of atoms (geometry) is the same as the electronic arrangement. It is tetrahedral. 11 1.26. (a) The carbon atom highlighted (below) has three bonds and no lone pairs (steric number = 3) and is therefore sp2 hybridized and trigonal planar. H H H C C H H O C N C O H C H C H H H H Each of the other carbon atoms has four bonds (steric number = 4) and is therefore tetrahedral. The nitrogen atom has three bonds and one lone pair (steric number = 4), so the electronic arrangement is tetrahedral. But one corner of the tetrahedron is occupied by a lone pair, so the arrangement of atoms is trigonal pyramidal. The oxygen atom (of the OH group) has two bonds and two lone pairs (steric number = 4), so the electronic arrangement is tetrahedral. But two corners of the tetrahedron are occupied by lone pairs, so the arrangement of atoms is bent. The oxygen atom of the C=O group has one  bond and two lone pairs (steric number = 3), so it is sp2 hybridized. (b) Each of the highlighted carbon atoms has three bonds and no lone pairs (steric number = 3) and is therefore trigonal planar. Each of the other carbon atoms (not highlighted) has four bonds (steric number = 4), with tetrahedral geometry. The highlighted nitrogen atom has two bonds and one lone pair (steric number = 3), so the electronic arrangement is trigonal planar. But there is one lone pair, so the arrangement of atoms is bent. The other nitrogen atom (not highlighted) has three bonds and one lone pair (steric number = 4), so the electronic arrangement will be tetrahedral. One corner of the tetrahedron is occupied by a lone pair, so the arrangement of atoms is trigonal pyramidal. The oxygen atom has two bonds and two lone pairs (steric number = 4), so the electronic arrangement will be tetrahedral. Two corners of the tetrahedron are occupied by lone pairs, so the arrangement of atoms is bent.   12 CHAPTER 1 (c) Each of the carbon atoms has three bonds and no lone pairs (steric number = 3) and is therefore trigonal planar: 1.27. The carbon atom of the carbocation has three bonds and no lone pairs (steric number = 3), and is therefore trigonal planar. The carbon atom of the carbanion has three bonds and one lone pair (steric number = 4), and is therefore trigonal pyramidal (the electronic arrangement is tetrahedral, but one corner of the tetrahedron is occupied by a lone pair, giving a trigonal pyramidal arrangement of atoms). 1.28. Every carbon atom in benzene has three bonds and no lone pairs. With a steric number of 3, each of these carbon atoms is sp2 hybridized and trigonal planar. Therefore, the entire molecule is planar (all of the atoms in this molecule occupy the same plane). 1.29. (a) This compound has three C–Cl bonds, each of which exhibits a dipole moment. To determine if these dipole moments cancel each other, we must identify the molecular geometry. The central carbon atom has four bonds so it is sp3 hybridized, with tetrahedral geometry. As such, the three C–Cl bonds do not lie in the same plane, and they do not completely cancel each other out. There is a net molecular dipole moment, as shown: other. There is a net molecular dipole moment, as shown: (d) The central carbon atom has four bonds (steric number = 4) and is therefore sp3 hybridized, with tetrahedral geometry. There are individual dipole moments associated with each of the C–Cl bonds and each of the C–Br bonds. If all four dipole moments had the same magnitude, then we would expect them to completely cancel each other to give no molecular dipole moment (as in the case of CCl4). However, the dipole moments for the C–Cl bonds are larger than the dipole moments of the C–Br bonds, and as such, there is a net molecular dipole moment, shown here: (e) The oxygen atom has two bonds and two lone pairs (steric number = 4), so it is sp3 hybridized (electronically tetrahedral), with bent geometry (because two corners of the tetrahedron are occupied by lone pairs). As such, the dipole moments associated with the C–O bonds do not fully cancel each other. There is a net molecular dipole moment, as shown: H H H C C H O H C C H H H (f) There are individual dipole moments associated with each C–O bond (just as we saw in the solution to 1.29e), but in this case, they fully cancel each other to give no net molecular dipole moment. (b) The oxygen atom has two bonds and two lone pairs (steric number = 4), so it is sp3 hybridized (electronically tetrahedral), with bent geometry (because two corners of the tetrahedron are occupied by lone pairs). As such, the dipole moments associated with the C–O bonds do not fully cancel each other. There is a net molecular dipole moment, as shown: (c) The nitrogen atom has three bonds and one lone pair (steric number = 4), so it is sp3 hybridized (electronically tetrahedral), with trigonal pyramidal geometry (because one corner of the tetrahedron is occupied by a lone pair). As such, the dipole moments associated with the N–H bonds do not fully cancel each (g) Each C=O bond has a strong dipole moment, and they do not fully cancel each other because they are not pointing in opposite directions. As such, there will be a net molecular dipole moment, as shown here: (h) Each C=O bond has a strong dipole moment, and in this case, they are pointing in opposite directions. As such, they fully cancel each other, giving no net molecular dipole moment.